PHY2049 General Physics II ELECTRICITY, MAGNETISM & LIGHT CHAPTER 21. Chapter 21 THE ELECTRIC FIELD I DISCRETE CHARGE DISTRIBUTIONS.

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1 PHY2049 General Physics II ELECTRICITY, MAGNETISM & LIGHT Chapter 21 The electric field: I discrete charge distributions Chapter 22 The electric field: II continuous charge distributions Chapter 23 Electric potential Chapter 24 Electrostatic energy and capacitance Chapter 25 Electric current and directcurrent circuits Chapter 26 The magnetic field Chapter 27 Source of the magnetic field Chapter 28 Magnetic induction Chapter 29 Alternatingcurrent circuits Chapters 30/31 Electromagnetic waves and the properties of light Chapter 32 Optical images CHAPTER 21 THE ELECTRIC FIELD I DISCRETE CHARGE DISTRIBUTIONS Electric charges Conductors and Insulators Coulomb s Law Electric field Electric field lines Point charges in an electric field Electric dipoles

2 Rubber rod rubbed with fur REPELS Rubber rod rubbed with fur Rubber rod rubbed with fur REPELS Wax rod rubbed with fur EFA02VD1.MOV Glass rod rubbed with silk ATTRACTS Rubber rod rubbed with fur Rubber and wax rods rubbed with fur. Glass rod rubbed with silk. Two types of charges: Like charges REPEL Unlike charges ATTRACT positive & negative (Benjamin Franklin ( ) Glass rod/silk Glass rod ( ve) F Rubber F Rubber Glass F F Rubber Rubber rod/fur Wax rod/fur Silk ( ve) Rubber rod ( ve) Fur ( ve) Wax rod ( ve) Fur ( ve)

3 Electrostatic series Glass Mica Wool Quartz Fur Silk Felt Cotton Wood Cork Amber Rubber Sulfur Positive Electric charge is quantized... Each individual charge e Total charge Q ne where n is an integer and e is the smallest, indivisible charge the charge of one electron. When in contact, each material becomes more positive than the ones below it. Example: glass rubbed with silk glass positive silk negative Typical values of n Charge is measured in Coulombs (later). e = C, Q ~ C.

4 DISCUSSION PROBLEM [21.1] We are used to thinking that various quantities like time, speed, length, vary smoothly, i.e., continuously. Can you think of several examples of quantized quantities, i.e., those that do not change continuously but from some value N to N ±1? Normally, in an insulator, like a piece of paper, the molecules are orientated randomly. Furthermore, there are the same numbers of positive and negative charges in each molecule and so, overall, the object is electrically NEUTRAL.

5 Fur electrons Plastic rod Plastic has a greater affinity for electrons than fur. So, if a plastic rod is rubbed with fur, the rod becomes negatively charged while the fur becomes positively charged. electrons Silk Glass rod Silk has a greater affinity for electrons than glass. So, if a glass rod is rubbed with silk, the rod becomes positively charged while the silk becomes negatively charged. Charged rod Attractive force polarized If a charged rod is brought near to a piece of paper, the charges in the molecules shift slightly so that each molecule has a slightly more positive end which in this case is attracted towards the rod and a slightly more negative end. However, the molecules do not move; they become polarized. normal

6 Comparison between insulators and conductors. Charging a conductor: 8 net charge: 0 8 Introduce charges in an object. In an uncharged conductor, like a metal sphere, there are equal numbers of positive and negative charges and so there is no net charge: the object is electrically NEUTRAL. If the object is an insulator, the charges remain localized. If the object is a conductor, the charges repel each other and rapidly disperse. Note: dry air insulator moist air conductor Therefore, conductors cannot be given a permanent charge by the same method as insulators as the charges can move away from each other. Conductors can be charged by another method... Negative charges can move freely in a conductor so when a negatively charged rod is brought near, negative charges tend to move away from the rod and cause the object to become polarized. The positive charges (ions) remain fixed. Note: charges are not created, they are REDISTRIBUTED

7 Charging by induction Wall outlet Ground (or Earth) connectors grounding symbol Grounding terminal (usually green) Grounding rod 8 4 net charge: 4 Ground A conductor can be given a permanent ve charge by removing some of the ve charges. The latter can be carried away through a wire attached to the ground or by touching the object with a finger... called grounding. The ground connectors of a wall outlet are connected to a long metal spike (called a grounding rod) driven into the ground.

8 Conceptual question: A B A B A B Question 21.1: Two, uncharged, insulated conducting spheres A and B are in contact. A positively charged rod is brought near A, but does not touch A. A and B are then separated while the charged rod remains near A. After the rod is removed, what are the charges on A and B? A Sphere A will be negatively charged and sphere B will be positively charged. This is another way of charging a conductor, instead of by induction. B

9 Q 1 Q 2 Coulomb s Law. Charles Augustin de Coulomb ( ) q 2 q 1 r r Q 1 Q 2 2 When two charged conducting spheres with the same radii but different charges touch, the charges redistribute so the charges are the same on each sphere. This is only true if the radii of the spheres are the same. In chapter 23 we deal with the case when the radii are different. Q 1 Q 2 2 F = k q 1 q 2 r 2 F = k q 1q 2 r 2 ˆ r If q 1 and q 2 are of the same sign, F > 0 REPULSIVE If q 1 and q 2 are of opposite sign, F < 0 q 1 q 2 F 21 q 1 q 2 F 21 F 12 F 21 = F 12 ATTRACTIVE F 12

10 Then: F = k q 1q 2 r 2 Charges q 1 and q 2 Coulombs (C) Distance r meters (m) Force Newtons (N) k = Nm 2 C 2 ( ) = 1 4πε, where ε = C 2 Nm 2, is called the permittivity of free space (i.e., vacuum). What if the charges are in a medium?... replace ε ε = κε, where κ is called the dielectric constant. Air κ Oil 2.2 Methyl Alcohol 33.6 Water 80.9 k k κ, so the force is reduced. Normally, for air we take κ = 1. Question 21.2: (a) In an atom of hydrogen the electron and proton are separated by an average distance of m. What is the magnitude of the mutual electrostatic force of attraction between the two? (b) Calculate the ratio of the electric force to the gravitational force between the two particles.

11 (a) m q 1 q 2 Proton and an electron a distance m apart. q 1 = C and q 2 = C F E = k q 1q 2 r 2 = ( ) ( ) ( ) 2 See... the gravitational force is very weak compared with the electric force. = N (attractive). (b) F G = G m em p r 2 m p = kg m e = kg = ( ) ( ) ( ) 2 = N (always attractive)... then how come gravity is the dominant force between objects like planets and stars? F E F G = =

12 Ah well It tells us that there must be an upper limit on the amount of charge on a star or planet. Question 21.3: The charges shown below are all point charges. Which configuration produces the largest repulsive force? M 1,Q R 1 M 2,Q 2 q 2q q q r r 2 2q 2q r 2 F E = k Q 1Q 2 R 2 F G = G M 1M 2 R 2 Since F E << F G, then kq 1 Q 2 << GM 1 M 2, i.e., Q << G k M or Q << M. For the Earth: M kg A B C 2q 2q 2q 2q 1.5r 2r D E Q << C (Actual value... ~10 6 C)

13 Answer B. q 2q r A q q r 2B 2q 2q r 2 C 2q 2q 1.5r D 2q 2q 2r E F = k 2q2 r 2 F = k q q r 2 = 2 k q2 r 2 ( ) 2 = 4 k q Force is attractive, not repulsive 2 r 2 2q 2q q2 F = k 2 = 1.78 k (1.5r) r 2 2q 2q F = k (2r) 2 = k q2 r 2 When there are several electric forces acting we must take the VECTOR SUM, viz: 1 (Fixed) 3 2 (Fixed) F 13 F 23 F R The resultant F R = F 23 F 13. In this example if 1 and 2 are fixed... then 3 will experience an initial acceleration in the direction of given by Newton s 2nd Law a F = R. m3 F R

14 Question 21.4: Three point charges, each of magnitude 3.00nC, are at separate corners of a square whose sides are 5.00cm long. The two charges on opposite corners are positive and the third charge is negative. Find the force by these charges on a fourth point charge q 4 = 3.00nC at the remaining corner C C 3 5 cm 2 5 cm C F " F 24 F 14 F 14 = k ( ) ( ) ˆ ( ) 2 i = N î. F 24 = k ( ) ( ) ( ) 2 ˆ j = N ĵ. F 34,x = F 34 cos135 " ˆ i = F 34 cos45 " ˆ i. F 34,y = F 34 sin135 " ˆ j = F 34 sin 45 " ˆ j. ĵ î We will need the distance between 3 and 4 : = ( ) 2 ( ) 2 = m.

15 C C 3 5 cm 2 5 cm C F " F 24 F 14 ĵ î The Electric Field: Q 1 r The electric force between these two charges is: F = k Q 1Q 2 ˆ r. r 2 Q 2 and F 34,x = k ( ) ( ) ( ) ˆ i = N î F 34,y = k ( ) ( ) ( ) ˆ j = N ĵ F R = F 14 F 24 ( F 34,x F 34,y ) = ˆ i ĵ ( ˆ i ˆ j ) = ( ˆ i ˆ j ) N We say it is the electric field produced by Q 1 that Q 2 feels and produces the force on Q 2 (and vice versa). The electric field of Q 1 at r (i.e., at the position of Q 2 ) is defined as: E ( r ) = Force on Q 2 F = = k Q 1 Q 2 Q 2 r 2 ˆ r Note also, the force on Q 2 is F = Q 2 E. So, we can find the electric field at r, E ( r ), due to Q 1 by using a test charge q " (of known charge) in place of Q 2 and measuring F

16 cf: the Gravitational Field: M The gravitational force that the apple experiences due to the presence of the Earth is its weight: F = m g. We say it is the gravitational field produced by M (the Earth) that m (the apple) feels and produces the force on the apple. Using the same definition we used for the electric field, the gravitational field of M at r (i.e., the position of m) is: Earth Force on m m = m g m = g so, g is equivalent to E. Note also that: F G = m g and F E = q E. However, F G is always attractive whereas F E can be attractive OR repulsive. r apple m So, we can determine both the magnitude and direction of E by using a test charge q " to find F, since E = F q " However... we must assume that our test charge does not affect the charge Q, i.e., (a) with no test charge, the charge is distributed uniformly over the sphere. (b) if q " is large, it may affect the way the charge Q is distributed. Thus, the physically correct definition of the electric field is: q " E = Limit q" 0 F q" But we do not need to be this specific... we always assume that a test charge q " is very small. Furthermore, we always assume that the test charge q " is positive, so that E l F.

17 Electric field lines An electric field can be shown as a set of lines that are parallel to the electric field vector at any point in space. These lines have the following properties: The electric field vector E is tangent to the electric field line at every point. Electric field lines are directed away from a ve charge and towards a ve charge. The number of lines per unit area perpendicular to the lines is proportional to the magnitude of the electric field in that region. Electric field lines cannot cross. A B E For example, the density of lines through A is greater than the density through B. Therefore, the electric field is larger at A than at B. Electric field lines for a single isolated positive charge: F a > F b Q q " F E (r) = = 1 k Qq " q " q " r 2 ˆ r = k Q r 2 ˆ r. The field is radial and its magnitude depends only on r (= r). For positive Q, the field lines are directed outward, since F ( = q " E ) is directed radially outward and E l F, when q " is positive. The number of lines leaving the charge is proportional to Q, i.e., a charge of 2Q would have twice as many field lines. a b

18 The electric field variation along the xdirection for a charge Q at the origin: E (x) = k Q x 2 î. The direction of the electric field is determined by the direction of the force experienced by a positive test charge. E x EFM10VD1.MOV Q î x Question 21.5: Why does the girl s hair stand on end when she puts her hand on a Van der Graaf generator? NOTE: She is standing on an insulated table. ve direction x < 0 q " q " ve direction x > 0

19 Electric field due to two equal but opposite charges: 1 2 Q Q when her hair stands on end. The girl s body becomes charged... so does her head and her hair. The strands of hair repel each other... the most stable arrangement is Because they are charged, the individual hairs follow closely the lines of the electric field produced by her head... they act as test charges. E 2b E 1b b E 2a E a E b E 1a The electric field is a vector and so the electric field at any point due to a number of charges is the vector sum of the individual electric fields: i.e., E ( r ) = i E i ( r ). Note in this example: E 1a = E 2a but E 1b > E 2b. a

20 Electric field lines for two equal but opposite charges: c The electric field variation along the xdirection for two charges ±Q, at x = ±a. The direction of the electric field is determined by the direction of the net force experienced by a positive test charge. a Q Q F a > F b > F c E a > E b > E c E x b The number of lines leaving or entering a charge depends on the magnitude of the charge (we will prove this later). Since the magnitudes of the charges are the same, the same number of lines leave the ve charge (source) as enter the ve charge (sink). ve x < a ve a < x < 0 ve a > x > 0 ve x > a î x

21 Question 21.6: Suppose we have two oppositely charged objects and we separate them farther. Has the energy of the system of charges changed or is it unchanged? Explain your answer. a The energy of the Q 1 Q 2 system of charges has b increased. How do I know that? Because I have to do positive work to separate the charges farther. Assume initially the charges are a distance a apart and I move them to a distance b apart. Then the work done is b W = F d x b = k Q 1Q 2 dx a = kq 1 Q 2 1 x a b a x 2 1 = kq 1 Q 2 a 1 > 0. b Where does the energy go?... into the electric field After all an electric field can do work (by making charges move), so it must contain (potential) energy. In another chapter we will calculate the energy of an electric field. Think... what about gravitational and magnetic fields... do they contain energy?

22 Electric field lines for two positive charges with the same magnitude: A Q Q E 2 E E 1 A Q Q F 2 F 1 F Question 21.7: Two point charges, q 1 and q 2 each with a charge of 6.00nC, are placed on the yaxis at y 1 = 3.00cm and y 2 = 3.00cm, respectively. (a) What are the magnitude and direction of the electric field on the xaxis at x = 4.00cm? Since the magnitudes of the charges is the same, the same number of lines leave each charge. (b) What is the force exerted on a third charge q 3 = 2.00nC placed on the xaxis at x = 4.00cm? What would be the force on a test charge at the point A, midway between the two charges?

23 1 3 cm 3 cm 2 4 cm 5 cm 5 cm E 2 θ E R E 1 (a) The resultant electric field: E R = E 1 E 2. Note that by symmetry: E 1,y E 2,y = 0. E 1,x = k ( ) 2 cosθ ˆ i = ˆ i N/C E 2,x = k ( ) 2 cosθ ˆ i = ˆ i N/C E R = ˆ i N/C. θ ĵ î The electric field along the xdirection variation for two charges Q, at x = ±a. The direction of the electric field is determined by the direction of the net force experienced by a positive test charge. E x î x (b) Force on charge q 3 = 2 nc: F = q 3 E R = ˆ i = ˆ i N ve x < a ve a < x < 0 ve x > a ve a > x > 0

24 Q 1 Q Q 1 5 Q Question 21.8: (Review question.) Shown here are the electric field lines near two unlike charges of different magnitudes. (a) What are the signs of the charge on Q 1 and Q 2? (b) What is the ratio Q 1 Q2? (c) Some of the electric field lines from Q 1 end on Q 2. Where do the electric field lines that do not end on Q 2 go? Do they just stop? (a) The electric field lines are outward from Q 1 (so it s ve) and inward towards Q 2 (so it s ve). (b) The number of electric field lines going into or out of a charge depends on the magnitude of the charge. The ratio of the number of lines leaving Q 1 to the number of lines going into Q 2 is: 16 8 = Q 1 Q2 = 2, i.e., Q 1 = 2Q 2.

25 (c) No, they don t stop Electric field lines always begin on positive charge (the source) and end on negative charge (the sink). Therefore, the extra lines must end up on nearby negative charges, such as on you, or on the table, or on the walls In this example, a rubber rod ( ve) has been rubbed by fur ( ve). Note that many of the electric field lines end on the walls of the room Q Q F = q " E E Electric field for two oppositely charged plates (with total charge ±Q). Notice that the electric field lines are parallel and uniformly spaced except at the edges of the plates. So, the electric field between the plates is constant, which means that the force exerted on a charge is also constant between the plates. We will derive the expression for the electric field between two charged plates in the next chapter.

26 35 " v " E = E yˆ j 1.50 cm ˆ j y ˆ i x Question 21.9: A charged particle leaves the origin with a speed of m/s at an angle of 35 above the x axis. A uniform electric field, E " = E ˆ j, exists throughout the region. Find E such that the particle will cross the xaxis at x = 1.50cm if the particle is (a) an electron, and (b) is a proton. (a) To deflect an electron downward, the field must be directed upward. We have... v "x = cos35 " ˆ i = ˆ i m/s. v "y = sin 35 " ˆ j = ˆ j m/s. The force on the electron is F = q E = ee y ĵ, i.e., it is directed only in the y direction; there is no force (i.e., no acceleration) in the x direction, so v "x is constant. (Does this remind you of something we ve studied before?). Time connects the x and y motions. The time of flight from x = 0 to x = 1.50 cm is: t = Range speed = v "x = s.

27 35 " v " E = E yˆ j 1.50 cm In the ydirection, we have: a y = F m = ee y m and v y = v y a y t. But, when the electron reaches the bottom plate, i.e., E y = 2 v y m et t = 2v y a y v y = v y, ˆ j y ˆ i = 2 v y m ee y = s. = m/s kg C s = N/C. x 35 " v " E = E yˆ j 1.50 cm (b) To deflect a proton downward, the field must be directed downward. We have again... v "x = cos35 " ˆ i = ˆ i m/s v "y = sin 35 " ˆ j = ˆ j m/s The force on the proton is F = q E = ee y ĵ, i.e., it is directed only in the y direction; there is no force (i.e., no acceleration) in the x direction, so again v "x is constant. Time connects the x and y motions. The time of flight from x = 0 to x = 1.50 cm is: t = Range speed which is the same as before. ˆ j = v "x = s, y ˆ i x

28 35 " v " E = E yˆ j 1.50 cm In the ydirection, we have: a y = F m = ee y m and v y = v y a y t. But, when the proton reaches the bottom plate, i.e., t = 2v y a y E y = 2 v y m et v y = v y, ˆ j = 2 v y m ee y = s. = m/s kg C s = N/C. Note that E y m. Since m p > m e, the electric field is greater in the case of the proton by the ratio m p me. y ˆ i x Electric dipoles... The electric dipole moment is defined as: p = q d where d is the displacement from q to q. In a uniform field, the forces F 1 and F 2 are equal but opposite and so, they form a couple. The magnitude of the net torque on the dipole about the rotation axis is l τ = F 1 2 F l 2 = Fl = (qe)(d sin θ) = pesin θ, 2 where θ is the angle between the dipole moment and the field direction. In this case the dipole will rotate clockwise. l = dsin θ F 2 = q E q d θ q F 1 = q E NOTE: there is no net force acting on the dipole

29 Potential energy of a dipole in a field... p The work done by the field to θ rotate the dipole through dθ is: E dw = τdθ = pesin θdθ. Therefore, the change in potential energy is: So, the potential energy of the dipole is: U = du du = dw = pesin θdθ. = pe sin θdθ = pecosθ = p E When θ = 0, U has its minimum value (stable equilibrium). When θ = 180 ", U has its maximum value (unstable equilibrium). U o 360 o θ H H O The water molecule is a POLAR MOLECULE with p = C m. That fact is the basis of the operation of a microwave oven. The oscillating electric field within a microwave oven produces an oscillating torque on the water molecules that causes them to flip backandforth, producing heat. The heating effect is maximal at microwave frequencies (about 2.45 billion Hz). The torque acts to align the dipole with p parallel to E.

30 If a molecule is placed in an nonuniform electric field (say, due a single isolated positive charge), the molecule may become polarized, i.e., the positive and negative charge distort slightly forming a dipole. Since the negative charge is closer to the point charge than the positive charge... F 1 F 2... there is a net attraction. F 1 > F 2 Why? Question 21.10: What difference would it make if the central charge (shown in the previous slide) were negative instead of positive? A: The molecule would be repelled from the central charge. B: The molecule would remain stationary C: The molecule would be attracted to the central charge.

31 F 1 F 2 If the charge were negative, the electric field lines would be directed inward. The molecule would again be polarized but now, the positive end would be closer to the central, negative charge. Again F 1 > F 2 so the molecule would still be attracted to the charge. So, it makes no difference what the sign of the charge is, there is always attraction Therefore, the answer is C. The electric field of a positively charged plastic comb produces an attractive force on the molecules in a stream of water. H H O

32 Question 21.11: What would happen to a neutral (but polarizable) object if it were placed in a uniform field? A: Nothing, it would remain stationary. B: It would move to the left. C: It would would to the right. D: It would rotate clockwise. E: It would rotate counterclockwise. The electric field exerts forces on the charges in the object so the molecules would become polarized, with the ve charges moving in the opposite direction to the field, i.e., to the left. The object becomes a dipole and would experience a torque that tends to turn it clockwise. The answer is D... it would rotate clockwise. Note: because the sizes of the ve and ve charges are the same, there is no net force on the object, so there is no translational motion, only rotational motion